According to rounding rules for addition the sum of 27.1, 34.538, and 37.68 is

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Why is it so hard to dispose of nuclear waste

Anarel [89]

It is highly hazardous, it is radioactive

8 0

3 years ago

NaOH + HCl → NaCl + H2O

bearhunter [10]

Answer:

0.00496 g H2O

Explanation:

7 0

3 years ago

{Science Question} -Photosynthesis-

MAVERICK [17]

Look I am not even close to a genius in this type of stuff but the leaves help collect energy and sunlight for the plant to live off of during the winter. It basically needs them for processing and helping them with their food. Also food is stored in every pine needles mitochondrion or the powerhouse of their cell structure. In the mitochondrion their is things like chloropasts. These are like the energy or food for the cell. So when the tree cant draw nutrients from its roots or soil anymore all the excess things stored in the pine needles will help the tree survive.

3 0

3 years ago

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The dimensions of a cupboard are 1.31 m wide

Mnenie [13.5K]

Answer:

The volume of cupboard is 2.0043 m³.

Explanation:

Given data:

width of cupboard = 1.31 m

length of cupboard = 0.9 m

height of cupboard = 1.70 m

Volume = ?

Solution:

Volume = length × width × height

Volume = 0.9 m × 1.31 m × 1.70 m

Volume = 2.0043 m³

The volume of cupboard is 2.0043 m³.

4 0

3 years ago

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